Where the estimate stops being a bound

A sum that is exact, and the estimate that is not made a claim it was careful about and did not test: that the bandwidth estimate built from open-circuit time constants is never optimistic. On every network that essay drew, it is. Three resistors and three capacitors in a chain give three real poles, the estimate comes out fourteen per cent low at equal values and converges upward as one time constant takes over, and at no point on the slider does it sit above the measured corner.

The essay also said, in its list of things it was not claiming, that the property fails for a resonant network and fails badly. That sentence was a piece of reasoning rather than a measurement, and reasoning is the thing that gets a number put on it here.

The number turns out to be exact, and there are two of them. Both are $\sqrt2$, and the more useful of the pair is not the one the argument was started for: the estimate’s worst point is at a quality factor of $1/\sqrt2$, which is the Butterworth value and is therefore the most commonly built second-order section in existence.

A section whose sum of time constants cannot move #

The network is the ordinary equal-resistor Sallen–Key low-pass: a kilohm from the input to node a, a second kilohm from a to node b, a capacitor from b to ground, a second capacitor from a to the output, and a unity-gain buffer from b to the output. Its quality factor is $\tfrac12\sqrt{C_1/C_2}$ and its resonance is $1/2\pi R\sqrt{C_1C_2}$, so sweeping the ratio of the two capacitors sweeps the quality factor while leaving everything else about the topology alone.

Its open-circuit time constants are unusual, and one of them is worth stopping on.

The feedback capacitor sees exactly zero resistance, and it is not a numerical accident. With the other capacitor removed, node b is connected to node a through a resistor carrying no current, so $v_b = v_a$; the buffer holds the output at $v_b$; and the two terminals of $C_1$ are at the same potential whatever is injected into them. A probe current into node a raises the output by exactly as much. The amplifier is doing what an amplifier does, which is to make a resistance look like something else, and here it makes two kilohms look like nothing.

So $\Sigma\tau = 2RC_2$ and nothing else. Substituting the definitions gives $\Sigma\tau = 1/(Q\omega_0)$ and therefore an estimate of $Q f_0$ — and since $R$ and $C_2$ are both held fixed while $C_1$ does the sweeping, the estimate is the same number at every setting of the slider. 7957.7 Hz, from one end of the axis to the other, while the thing it is estimating falls by a factor of thirteen.

That is the cleanest possible arrangement for the question being asked. One quantity is nailed down and the other walks past it.

The theorem does not care that the poles have gone complex #

Before the estimate is examined it is worth establishing that the exact half of the statement survives, because a reader who has watched the estimate fail may reasonably suspect the theorem of failing with it.

It does not. $\sum R_iC_i = a_1/a_0$ is a statement about a denominator polynomial and says nothing about where the roots are. It is checked here at thirty-six quality factors from 0.3 to 20 — real poles at one end, a pair with a quality factor of twenty at the other — against the ratio of the first two coefficients of a denominator recovered by sampling a determinant on a circle in the complex plane. The worst disagreement over the whole sweep is a part in ten million, which is the recovery’s tolerance and not the theorem’s, exactly as a sum that is exact found and for the same reason.

The measured corner is checked too, against the closed form $\omega_0\sqrt{k + \sqrt{k^2+1}}$ with $k = 1 - 1/2Q^2$, and they agree to a part in a hundred thousand at every point. Two routes to the truth and two routes to the estimate, and none of the four shares arithmetic with the others.

The two crossings, and both of them are root two #

Divide the estimate by the truth. With $k = 1 - 1/2Q^2$,

$$\frac{\text{estimate}}{\text{measured}} = \frac{Q}{\sqrt{k + \sqrt{k^2+1}}}$$

and everything that follows is a property of that one expression.

At $Q = 1/\sqrt2$ it equals exactly $1/\sqrt2$, and that is its minimum. Put $Q^2 = 1/2$ and $k$ is zero, so the denominator is $\sqrt{0+1} = 1$ and the ratio is $Q$ itself. The estimate is 29.29 per cent low — worse than anywhere else on the axis — at the one quality factor a designer is most likely to have chosen, because it is the maximally flat one. The figure golden-sections the minimum on the solved ratio rather than substituting, and lands on 0.70711.

At $Q = \sqrt2$ it equals exactly 1. Put $Q^2 = 2$ and $k = 3/4$, so $\sqrt{k+\sqrt{k^2+1}} = \sqrt{0.75+1.25} = \sqrt2 = Q$. The estimate is not merely close there; it is right, and it is right by a cancellation rather than by design. Bisected on the solved response — a corner measured on a nodal solve, inside a bisection on the ratio — the crossing comes out at 1.414213 against $\sqrt2 = 1.414214$.

Above it the estimate is optimistic and there is no bound at all. At a quality factor of ten it claims 7957.7 Hz for a section whose corner is 1234.3, which is 6.45 times the bandwidth there is. At twenty it is 12.9 times. The ratio grows without limit, because the estimate is frozen and the truth falls as $1/Q$.

Why the bound held on a chain and does not hold here #

The reason the open-circuit sum is conservative on a network with real poles is that $\sum(-1/p_i)$ weights every pole equally while the magnitude response is dominated by the slowest one. Summing the reciprocals always over-counts unless a single pole is doing all the work, so the estimated corner is always at or below the real one.

A complex pair breaks the second half of that sentence. Neither pole is slower than the other — they are conjugates and have the same magnitude — and the response near the corner is not falling off towards the slower of them but peaking, because the two contributions are adding with a phase between them. The magnitude at $\omega_0$ is $Q$ times what a single pole at $\omega_0$ would give, and the corner is pushed out past the resonance by a factor that grows with $Q$. The sum of time constants has no term that grows with $Q$; it has a term that shrinks with it, since $\Sigma\tau = 1/(Q\omega_0)$.

So the two quantities move in opposite directions in $Q$, which is why they cross exactly once and why the crossing is a clean number rather than a fitted one. The same shape of failure appears wherever a configuration is summarised by an average over its poles: the cancellation that leaves a tail is a settling time no pole location predicts because it is a residue rather than a rate, and a floor, or a line is a jitter total that two different spectra can share. The estimate goes as $Q\omega_0$ and the truth goes as $\omega_0\sqrt{k+\sqrt{k^2+1}}$, and $\omega_0$ divides out of the comparison entirely — which is the reason the answer is a pure quality factor with no frequency, no resistance and no capacitance in it.

What a designer is actually exposed to #

A quality factor of $\sqrt2$ sounds like a corner case until the filters are counted.

A Butterworth section is at 0.7071 and is safe — it is at the estimate’s worst point, but the error is in the conservative direction. A Bessel pair is at 0.5774, also safe. A fourth-order Butterworth has sections at 0.5412 and 1.3066, and the second of those is just under the crossing, which means a fourth-order maximally flat filter is the last order at which the shortcut is a bound. A sixth-order Butterworth has a section at 1.932. A 0.5 dB Chebyshev of fourth order has one at 2.94, and of sixth order one at 8.00. Every elliptic section of any useful order is well above $\sqrt2$.

So the rule is not “beware of resonant circuits”. It is: the open-circuit sum is a bound on amplifiers and is not a bound on filters, and the boundary between the two populations sits at fourth-order maximally flat. That is why the shortcut is taught in amplifier design and not in filter design, and it is a better reason than the one usually given, which is that filters are designed from tables and amplifiers are not.

It is also a reason to be careful about where the two populations meet. Three families, one corner sets three fourth-order responses of the same corner beside each other, and two of the three have a section above $\sqrt2$; what a steep skirt costs is the same trade seen from the stopband. An active filter is an amplifier with a resonant network round it, so a designer who reaches for the shortcut while building one is reaching for a tool that works on the amplifier and not on the thing the amplifier was put there to make.

The same section seen from the other direction is where the behaviour is written down: a pair of poles, a quality factor and a resonance, from which the corner follows exactly. That essay’s whole point is that the pair carries the behaviour, and this one is the price of summarising the pair in one number.

What the estimate is still good for at a high quality factor #

Not the corner. The decomposition, and it survives intact.

$\Sigma\tau$ is $2RC_2$ here and zero for the feedback capacitor, which says something a designer can act on: the section’s dominant time constant belongs to one of its two capacitors and the other one does not appear in it at all. A tolerance on $C_1$ moves the resonance and the quality factor and leaves $\Sigma\tau$ exactly where it was; a tolerance on $C_2$ moves all three. That asymmetry is real, it is what the direction a response is most sensitive to measures in general, and the sum finds it in one line of arithmetic on a resistor-only netlist.

It also says which quantity an amplifier’s finite bandwidth will damage first, which is the question the Q the amplifier decides answers in detail for this exact section. The term that is zero is zero because the buffer is perfect. Give the buffer a gain–bandwidth product and the resistance $C_1$ sees is no longer zero, the sum acquires a second term, and the size of that term is a direct measure of how far from ideal the buffer is at the frequency that matters. The shortcut turns into an instrument pointed at the amplifier rather than at the filter.

The failure is easier to see in the poles than in the response, so the last figure abandons the frequency axis entirely.

Four things this does not settle #

That the crossing is at $\sqrt2$ for every second-order response. It is for a low-pass whose numerator is a constant. A band-pass or a notch of the same quality factor has a different relationship between its poles and its −3 dB points, and the sum of open-circuit time constants knows nothing about numerators. The claim here is about one transfer function shape, measured across its whole quality-factor range.

That the zero-resistance term is a property of Sallen–Key sections. It is a property of unity gain. Give the amplifier a gain of $K$ and the resistance $C_1$ sees becomes $R_1(1-K)$, which is negative for $K > 1$ — the mechanism that makes the section’s quality factor adjustable by gain, and which puts a negative term into a sum that was introduced as a sum of positive ones.

That 6.45 times is the worst it gets. It is the worst on this axis. For a large quality factor $k \to 1$, so the measured corner tends to $\omega_0\sqrt{1+\sqrt2} = 1.5538\,\omega_0$ while the estimate goes as $Q\omega_0$ — the ratio therefore grows in proportion to $Q$ and without limit. A section with a quality factor of a hundred is out by a factor of sixty-four.

That the theorem is ever in doubt. Every figure in this essay claims it afresh at thirty-six quality factors, and the reason is that a reader watching an estimate fail this badly is entitled to wonder whether the identity underneath it is failing too.

The identity at thirty-six quality factors, and the crossing bisected #

The identity is checked at every quality factor on the sweep, against a denominator recovered from complex determinants, and comes out at a part in ten million at the worst of thirty-six points.

The measured corner is checked against its closed form at every one of them, to a part in a hundred thousand, so the falling curve is not being compared with itself.

The crossing is bisected on the solved response, not substituted, and checked against $\sqrt2$ to five parts in a million — the tolerance being the measurement’s, since the ratio’s slope through the crossing is 0.83 per decade of quality factor and the corner is itself the output of a bisection on a magnitude.

The worst point is checked as a value rather than as a location. A minimum cannot be located to the precision its value is known to — a quadratic bottom turns a part in $10^{10}$ of ratio into a part in $10^5$ of quality factor — so the sharp claim is that the ratio at $1/\sqrt2$ is $1/\sqrt2$, and the minimum is established by comparison with its neighbours and by a search that lands there.

An exact bound, and the exact place it stops #

Three essays on these essays have now measured the same two-layer object: an identity that is exactly a ratio of two polynomial coefficients, and a claim built on top of it that the ratio is a corner frequency.

The identity has never failed. It survives real poles, complex poles, a quality factor of twenty, a term that is exactly zero because an amplifier made it so, and being read from either end of the polynomial. It is arithmetic about a determinant and it has no opinion about circuits at all, which is the property one solve, read four ways relies on every time it takes four answers off one factorisation.

The claim on top has now failed in three different ways, each of them at a stated place. It is fourteen per cent low on a chain of three real poles and converges upward as one dominates. It is sixteen per cent high at the other end of the same polynomial and converges downward. And it changes sign at a quality factor of $\sqrt2$, above which it is not a bound in either direction.

What makes the third failure different from the first two is that the first two are quantitative — the estimate is wrong by a percentage that shrinks under a stated condition — and this one is qualitative. A bound that errs in one direction is a bound and can be designed against. A bound that errs in both is an estimate, and the difference decides whether a design can be signed off on it. The whole value of the fourteen per cent in a sum that is exact was that it was one-sided, and $\sqrt2$ is where the one-sidedness ends.

Which is the more useful form of the answer than “it does not work for resonant circuits”. That sentence is true and cannot be acted on. A quality factor of $\sqrt2$ can be looked up in a table of filter sections in about ten seconds, and it puts every Butterworth section up to fourth order on one side of the line and every Chebyshev section above second order on the other.

Still open: the sum with the amplifier’s own bandwidth in it, and the bound for a real pair #

The term that stops being zero. The feedback capacitor sees no resistance because the buffer is perfect. A buffer with a gain–bandwidth product holds its output at the input only up to a frequency, so the resistance is zero at direct current and something else at the corner — and the open-circuit construction is a direct-current measurement. Whether the sum computed with a real amplifier is closer to the corner or further from it, and whether the extra term has the right sign to repair the crossing, is a measurement on a netlist with one more element in it.

Where the crossing goes with a gain above one. A Sallen–Key section with a gain of $K$ has a negative open-circuit term, so its sum is smaller than the unity-gain section’s at the same quality factor and its estimate is correspondingly higher. That should move the crossing down — possibly below the Butterworth value, which would put every second-order section built for gain on the wrong side of it. Solved against $K$, it would say whether the safe population is the one this essay described or a smaller one.

And the bound for a pair that is not alone. Everything here is one second-order section. A fourth-order filter is two of them, one below $\sqrt2$ and one above on a Butterworth, and the sum over all four capacitors is a single number in which one conservative term and one optimistic term are added together. Whether the combination is a bound, and at what order it stops being one, is the question a filter designer would actually ask.